The derivative of arccossec(x) is:

Question

jpsmarinho · Answer

is that correct?

anonymous · Answer

$$\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$

anonymous · Answer

derivative of cosecant is $-\csc(x)\cot(x)$ if i remember correctly, so your answer is 
$$-\frac{1}{\csc(\csc^{-1}(x))\cot(\csc^{-1}(x))}$$ but you can clean up the denominator and leave no trace of a trig function

anonymous · Answer

now that i look at your answer, it is right, but you need to replace $y$ by $\csc^{-1}(x)$ to complete it

jpsmarinho · Answer

But, y is equals to arccossec(x) right?

anonymous · Answer

right

anonymous · Answer

and further 
$\csc(\csc^{-1}(x))=x$ and also $\cot(\csc^{-1}(x))=\sqrt{x^2-1}$

jpsmarinho · Answer

Aah when you told csc^-1(x) , I think that you wrote 1/csc(x). Thanks!

anonymous · Answer

yeah those  mean inverse functions

anonymous · Answer

so your "final answer" is $-\frac{1}{x\sqrt{x^2-1}}$

jpsmarinho · Answer

Wait, @satellite73 , please. How I discover that? cot(arccsc(x)) is equals to (x^2 - 1)^(1/2)? It's something about the identities sec(x) squared = tan(x) squared + 1?